If r + s > 2t, is r > t ?
(1) t > s
(2) r > s
Is there a simple method of solving such problems?
Using S1: If t > s, then r + t > r + s, and using the information in the question, r + t > 2t, or r > t.
Using S2: If r > s, then r + r > r + s, and using the information in the question, 2r > 2t, or r > t.
Alternatively you might see that the question is really about averages. We can rewrite the information in the question as (r+s)/2 > t. That is, we know that the average of r and s is greater than t. Now, we're just averaging two letters here, r and s. If they're different, one of them must be 'above average' (and therefore greater than t), and one must be 'below average' (and then possibly less than t, possibly not). Statement 1 tells us that s must be below average, so r must be above average; sufficient. Statement 2 tells us that s is smaller than r, so sufficient. D.
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